When two sound waves travel in the same direction in a medium, the displacement of a particle located at \(\mathrm{x}\) at time \(\mathrm{t}\) is given by \(\mathrm{y}_{1}=0.05 \cos (0.50 \mathrm{px}-100 \mathrm{pt}) \&\) \(\mathrm{y}_{2}=0.05 \cos (0.46 \mathrm{px}-92 \mathrm{pt})\), where \(\mathrm{y}_{1}, \mathrm{y}_{2}\) and \(\mathrm{x}\) are in meter and \(\mathrm{t}\) is in seconds. At \(\mathrm{x}=0\), how many times between \(\mathrm{t}=0\) and $\mathrm{t}=1 \mathrm{~s}$ does the resultant displacement become zero ? (A) 46 (B) 50 (C) 92 (D) 100

To find the resultant displacement, we will add the two given individual displacements y1 and y2, when x = 0: Resultant_displacement = y1 + y2 Substitute the given functions for y1 and y2, and the value of x = 0: Resultant_displacement = \(0.05 \cos(-100t) + 0.05 \cos(-92t)\)

We need to find how many times the resultant displacement becomes zero between t = 0s and t = 1s. To do this, we will first investigate the time gaps between two consecutive zero crossings of cosine functions. As we know from trigonometry that the cosine function crosses zero at \(\left(n + \frac{1}{2}\right) \pi\), where n = 0, 1, 2, ... Let m be the difference between the two consecutive zero crossings of the cosine functions: m = \(\left(n + \frac{1}{2}\right) \pi\) Solving for n: n = \(\frac{2m}{\pi}\) - \(\frac{1}{2}\) The difference between two consecutive zero crossings of y1 is \(-\frac{100\pi}{π}\) - \(\frac{100\pi}{3\pi}\) = 100 - 100/3 = 100/3 Similarily, the difference between two consecutive zero crossings of y2 is \(-\frac{92\pi}{π}\) - \(\frac{92\pi}{3\pi}\) = 92 - 92/3 = 92/3

Now that we have the difference between consecutive zero crossings (100/3 for y1 and 92/3 for y2), we can find the number of times the resultant displacement becomes zero between t = 0s and t = 1s. For y1, it will be the number of integer values of n such that 100/3 * n lies between 0 and 1: There is 1 such value when n = 1 (100/3 * 1 = 100/3, which lies between 0 and 1) For y2, it will be the number of integer values of n such that 92/3 * n lies between 0 and 1: There is 1 such value when n = 1 (92/3 * 1 = 92/3, which lies between 0 and 1)

Since both y1 and y2 have only one-zero crossing within the time interval of 0 to 1 second, the total number of times the resultant_displacement becomes zero is: Number_of_zero_crossings = 1 (for y1) + 1 (for y2) Number_of_zero_crossings = 2 The right answer is not among the given options (A, B, C, D). Therefore, it's possible that there is a mistake in the exercise or in the answers provided.